MATLAB: Create a block diagonal matrix - matlab

Question

I have a formula for creating a system of equations in matrix form.and wrote it like below:
for i=1:n+1
for j=1:n+1
t(i)=(1/2)*(1+cos(((2*(n-i)+3)*pi)/(2*(n+1))));
y(i,j)=t(i)^(j-1);
end
end
now what I have to do is creating a block diagonal matrix with rows of y separately in each block. i mean each row of y that contain powers of just t(i) should repeat three times in each block, then next row be the next block with three rows. like this:
T=[y(t1) 0 0;0 y(t1) 0;0 0 y(t1); ... ;y(t n+1) 0 0;0 y(tn+1) 0;0 0 y(tn+1)]

Answer 1

So since I can't comment in formated mode I try to interprete your question the following way:
So if
y =
1 2
3 4
Interpretation possibility a) is
T =
1 2 0 0 0 0 3 4 0 0 0 0
0 0 1 2 0 0 0 0 3 4 0 0
0 0 0 0 1 2 0 0 0 0 3 4
or this b)
T =
1 2 0 0 0 0 0 0 0 0 0 0
0 0 1 2 0 0 0 0 0 0 0 0
0 0 0 0 1 2 0 0 0 0 0 0
0 0 0 0 0 0 3 4 0 0 0 0
0 0 0 0 0 0 0 0 3 4 0 0
0 0 0 0 0 0 0 0 0 0 3 4
Solution
You first create one block i.e.
1 2 0 0 0 0
0 0 1 2 0 0
0 0 0 0 1 2
by calculating temp{1} = kron(I, y(1,:)), where I = eye(3). You then do the same operation for the second row: temp{2} = kron(I,y(2,:)) and so on. This procedure can be done easily in a for-loop:
I = eye(3);
temp = cell(1,size(y,1));
for i = 1:size(y,1)
temp{i} = kron(I, y(i,:));
end
Option a)
The result is now just: [temp{:}].
Option b)
Here the result is: blkdiag(temp{:})

Answer 2

you know, I had a problem for making a system of linear equations.maybe what I wrote above in calculating the y(i) just make it difficult for me.
for the example you said, I have this:
n=3
form equation above:
t1=0.0381
t2=0.3087
t3=0.6913
t4=0.9619
now,
y(t)=[1 t t^2 t^3 t^4]
I mean
y(t1)=[1.0000 0.0381 0.0014 0.0001]
y(t2)=[1.0000 0.3087 0.0953 0.0294]
y(t3)=[1.0000 0.6913 0.4780 0.3304]
y(t4)=[1.0000 0.9619 0.9253 0.8901]
but, since I couldn't calculate every y(i) seperately, wrote it in this form
y=[1.0000 0.0381 0.0014 0.0001
1.0000 0.3087 0.0953 0.0294
1.0000 0.6913 0.4780 0.3304
1.0000 0.9619 0.9253 0.8901]
now, should have two matrices. one:
s=[y(t) 0 0;0 y(t) 0;0 0 y(t)]
r=[s(t1) ... 0; ... ;0 ... s(tn+1)]
second:
d=[ 1 -1 1 -1
0 2 -8 18
0 0 8 -48
0 0 0 32]
k=y*d
u=[0 0 0;0 k 0;0 0 0]
a=[u(t1) ... 0; ... ;0 ... u(tn+1)]
and these matrices, r and a, should have 3(n+1) columns (in this example 12 columns).
I hope it could help